1. Electric Machine Magnetic CircuitBy
Dr. Shorouk Ossama

2. References
Stephen J. Chapman, “Electric Machinery Fundamentals”, 5th Edition, 2012.

3. An Electrical Machine is a device that can convert either mechanical energy to electrical energy, it is called a Generator or electrical energy to mechanical energy, it is called a Motor.
Almost all practical motors and generators convert energy from one to another through the action of a Magnetic Field.

4. The Transformer is an electrical device that is closely related to as electrical energy at another voltage level.
Electric motors in the: home run refrigerators, freezers, vacuum cleaners, blenders, air conditions, fans and mainly similar appliances. Of course, generators are necessary to supply the power used by all these motors.

5. Advantages For Using Electric Machines
Electric power is clean.
Easy to transmit over long distances and easy to control.
An electric motor does not require constant ventilation and fuel the way that an internal- combustion engine does.

6. Chapter (1): Magnetic Circuit
The Magnetic Fields are the fundamental mechanism by which energy is converted from one to another.
A current-carrying wire produces a magnetic field in the area around it.
A time-changing magnetic field induces a voltage in a coil of wire if it passes through that coil. (This is the basis of transformer action)
A current-carrying wire in the presence of a magnetic field has a force induced on it. (This is the basis of motor action)
A moving wire in the presence of a magnetic field has a voltage induced in it. (This is the basis of generator action)

7. Ampere’s Law
Where:
H is the magnetic field intensity produced by the current I
dl is a differential element of length along the path of integration
rectangular core with a winding of N turns of wire wrapped about one leg of the core.

8. The current passing within the path of integration I is the Ni.
The relationship between the magnetic field intensity H and the resulting magnetic flux density B produced within a material is:
Where: H is magnetic field intensity (Amp-Turn/m)
μ is magnetic permeability of material (H/m)
B is resulting magnetic flux density produced (webers per square meter Know as Tesla)

9. The permeability of free space is called μ0, and its value is:
μ0 = 4π x 10-7 H/m
The permeability of any other material compared to the permeability of free space is called its relative permeability:

10. The magnitude of the flux density is given by:
The total flux in the core:
Where: A is the cross-sectional area of the core

11. Magnetic fields can be visualized as lines of flux that form closed paths. Using a compass, we can determine the direction of the flux lines at any point. Note that the flux density vector B is tangent to the lines of flux.

12. right-hand rule

13. Electric Circuit
In a simple electric circuit: the voltage source V drives a current I around the circuit through a resistance R.
The relationship between
these quantities is given by
Ohm’s law: V = I R

14. Magnetic Circuit
In the magnetic circuit: The magneto motive force of the magnetic circuit is equal to the effective current flow applied to the core.
Where: F is the symbol for magneto
motive force (amperes-turns)

15. Analogue between a Dc Electric Circuit and a Magnetic Circuit

16. Reluctance of a Magnetic Path
The magnetomotive force (mmf) of N-turn current carrying coil is: The reluctance R of a magnetic path depends on the mean length l, the area A, and the permeability μ of the material. the relationship between magnetomotive force and flux is:

17. The equivalent reluctance of a number of reluctances in series is just the sum of the individual reluctances:
Req = R1 + R2 + R3 +…..
Similarly, reluctances in parallel combine according to the equation:
The reluctance calculation assumes a certain mean path length and cross sectional area (csa) of the core.

18. ferromagnetic cores
The core is just one block of ferromagnetic material (mean the core is composed of iron or certain other similar metals) with no corners, for practical ferromagnetic cores which have corners due to its design, this assumption is not accurate.
In ferromagnetic materials, the permeability varies with the amount of flux already in the material.

19. Example 1:
A ferromagnetic core in figure. The depth of the core (into the page) is 10 cm, and the other dimensions are shown in the figure. There is a 200 turn coil wrapped around the left side of the core. Assuming relative permeability μr of 2500, how much flux will be produced by a 1-Amp input current?

20. The core can be divided into two regions:
(1) the single thinner side and (2) the other three sides taken together. The magnetic circuit corresponding to this core is:

21. The mean path length of region (1) is 45, and the cross-sectional area is 10 x 10 cm = 100 cm2. Therefore, the reluctance in the first region is:
The mean path length of region (2) is 130 cm, and the cross-sectional area is 15 x 10 cm = 150 cm2. Therefore, the reluctance in the second region is:
( )

22. Therefore, the total reluctance in the core is:
Req = R1 + R2 = 14, ,600 =
41,900 A.truns/Wb
The total magnetomotive force (mmf) is:
F = Ni = 200 turns x 1 Amp = 200 A.truns
The total flux in the core is given by:

23. Example 2:
A ferromagnetic core whose mean path length is 40 cm. There is a small gap of 0.05 cm in the structure. The cross- sectional area of the core is 12 cm2, the relative permeability of the core is 4000, and the coil of wire on the core has 400 turns. Assume that fringing in the air gap increases the effective cross-sectional area of the air gap by 5 percent. find (a) the total reluctance of the flux path (iron plus air gap), and (b) the current required to produce a flux density of 0.5 T in the air gap.

24. The reluctance of the core is:
The reluctance of the air gap is: 1
1.05 x 12 = 12.6

25. Therefore, the total reluctance of the flux path is:
Req = Rc + Ra = 66, ,600 = 382,300 A.truns/Wb
(b) The total magnetomotive force
(F = Ni , F = ΦR) & the flux (Φ = BA), so the current is:

26. Example 3:
A simplified rotor and stator for a dc motor. The mean path length of the stator is 50 cm, and the cross-sectional area is 12 cm2, The mean path length of the rotor is 5 cm, and its cross-sectional area also may be assumed to be 12 cm2. Each air gap between the rotor and the stator is cm wide, and the cross-sectional area of each air gap; (including fringing) is 14 cm2. The iron of the core has a relative permeability of 2000, and there are 200 turns of wire on the core. If the current in the wire is adjusted to be 1 Amp, what will the resulting flux density in the air gap be?

27. The reluctance of the stator is:
The reluctance of the rotor is:

28. The reluctance of air gap is:
Therefore, the total reluctance of the flux path is:
Req = Rs + Ra1 + Rr + Ra = 166, , , ,000 = 751,000 A.truns/Wb
The total magnetomotive force applied to the core is:
F = N i = 200 x 1 = 200 A.turns

29. Therefore, the total flux in the core is:
Finally, the magnetic flux density in the motor’s air gap is:

30. Magnetic Behavior of Ferromagnetic Materials
The magnetic permeability was defined by the equation:
The permeability of ferromagnetic materials is very high, up to 6000 times the permeability of free space.
When the flux produced in the core is plotted versus the magnetomotive force the Saturation Curve or a Magnetizing Curve.

31. At first, a small increase in the magnetomotive force produces a huge increase in the resulting flux. After a certain point, though, further increases in the magnetomotive force produce relatively smaller increases in the flux; finally, an increase in the magnetomotive force produces almost no change at all.

32. Example 4:
Find the relative permeability of the typical ferromagnetic material whose Magnetizing Curve is shown in the figure at (a) H = 50, (b) H = 100, (c) H = 500 and (d) H = A.turns/m.

33. The permeability of a material is given by:
And the relative permeability is given by:
H = 50 A.turns/m
B = 0.25 T
μ = H/m
μr = 3980
H = 100 A.turns/m
B = 0.72 T
μ = H/m
μr = 5730
H = 500 A.turns/m
B = 1.4 T
μ = H/m
μr = 2230
H = 1000 A.turns/m
B = 1.51 T
μ = H/m
μr = 1200

34. Example 5:
A square magnetic core has a mean path length of 55 cm and a cross-sectional area of 150 cm2. A 200-turn coil of wire is wrapped around one leg of the core. The core is made of a material having the Magnetizing Curve shown in the figure.
How much current is required to produce Wb of flux in the core?
What is the core’s relative permeability at that current level?
What is its reluctance?

35. The required flux density in the core is:
From figure, the required magnetizing intensity is: H = 115 A.turns/m.
The magnetomotive force needed to produce this magnetizing intensity is:
F = Ni = H lc = 115 x 0.55 = A.turns

36. So the required current is:
The core’s permeability at this current is:
Therefore, the relative permeability is:
The reluctance of the core is:

37. Energy Losses In a Ferromagnetic Core
I increase
Path ab
Saturation Curve
I Falls
Path bcd
Hysteresis
Path dcdeb Hysteresis Loop

38. If a large magnetomotive force is first applied to the core and then removed, the flux path in the core will be abc. When the magnetomotive force is removed, the flux in the core does not go to Zero. Instead, a magnetic field is left in the core. This magnetic field is called the residual flux in the core as shown in the figure. To force the flux to Zero, an amount of magnetomotive force known as the coercive magnetomotive force FC must be applied to the core in the opposite direction.

39. Thanks
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